Problem Analysis #6

To find the probability that the point is back at $(2,1)$ after 3 seconds, we can analyze the transitions of the point under the four reflection operations:

1. Reflection in the origin ($O$): $(x,y) \to (-x, -y)$
2. Reflection in the $x$-axis ($X$): $(x,y) \to (x, -y)$
3. Reflection in the line $y=x$ ($A$): $(x,y) \to (y, x)$
4. Reflection in the line $y=-x$ ($B$): $(x,y) \to (-y, -x)$

Let the starting point be $S_1 = (2,1)$. The possible states the point can reach are:

• $S_1 = (2,1)$
• $S_2 = (2,-1)$
• $S_3 = (-2,1)$
• $S_4 = (-2,-1)$
• $S_5 = (1,2)$
• $S_6 = (1,-2)$
• $S_7 = (-1,2)$
• $S_8 = (-1,-2)$

At each step, one of the four operations is chosen with equal probability of $\frac{1}{4}$. After 3 steps, there are a total of $4^3 = 64$ possible sequences of operations.

We want to find the number of sequences of 3 operations $(Op_1, Op_2, Op_3)$ such that: $Op_3(Op_2(Op_1(S_1))) = S_1$

The only operations that can map a state to $S_1 = (2,1)$ in a single step are:

• $O$ applied to $S_4 = (-2,-1)$
• $X$ applied to $S_2 = (2,-1)$
• $A$ applied to $S_5 = (1,2)$
• $B$ applied to $S_8 = (-1,-2)$

Thus, the state after 2 steps, $Op_2(Op_1(S_1))$, must be in the set $\{S_2, S_4, S_5, S_8\}$.

Let's trace the first step $Op_1(S_1)$:

• If $Op_1 = O$, the state becomes $S_4$. From $S_4$:
  • $A(S_4) = S_8$, followed by $B(S_8) = S_1$ (Path: $O \to A \to B$)
  • $B(S_4) = S_5$, followed by $A(S_5) = S_1$ (Path: $O \to B \to A$)
• If $Op_1 = X$, the state becomes $S_2$. From $S_2$, no operation lands in $\{S_2, S_4, S_5, S_8\}$.
• If $Op_1 = A$, the state becomes $S_5$. From $S_5$:
  • $O(S_5) = S_8$, followed by $B(S_8) = S_1$ (Path: $A \to O \to B$)
  • $B(S_5) = S_4$, followed by $O(S_4) = S_1$ (Path: $A \to B \to O$)
• If $Op_1 = B$, the state becomes $S_8$. From $S_8$:
  • $O(S_8) = S_5$, followed by $A(S_5) = S_1$ (Path: $B \to O \to A$)
  • $A(S_8) = S_4$, followed by $O(S_4) = S_1$ (Path: $B \to A \to O$)

There are exactly $6$ successful paths of length 3:

1. $O \to A \to B$
2. $O \to B \to A$
3. $A \to O \to B$
4. $A \to B \to O$
5. $B \to O \to A$
6. $B \to A \to O$

Thus, the probability of returning to $(2,1)$ after 3 seconds is: $\frac{6}{64} = \frac{3}{32}$

This fraction is in lowest terms, so $p = 3$ and $q = 32$.

We find: $p + q = 3 + 32 = 35$

\boxed{35}